A particle moves along the curve $x^2y=36$ so that the $y$ -coordinate is changing at a constant rate of $-4$ units per minute. What is the rate of change (in units per minute) of the particle's $x$ -coordinate when the particle is at the point $(2,9)$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{3}$ (Choice B) B $-\dfrac{2}{9}$ (Choice C) C $-4$ (Choice D) D $\dfrac{4}{9}$
Background When working with motion along a curve, we should remember that this motion can also be represented by a position vector $(x,y)$. The main difference is that we're not given the equations for $x$ and $y$ in terms of time $t$, but only the relationship between $x$ and $y$ themselves. This, however, shouldn't prevent us from assuming the expressions for $x$ and $y$ in terms of $t$ exist. As always, $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$. Setting up the math We are given that $\dfrac{dy}{dt}=-4$ for any value of $t$. We are asked for the rate of change of the particle's $x$ -coordinate when the particle is at the point $(2,9)$. In other words, we are asked for the value of $\dfrac{dx}{dt}$ at the point $(2,9)$. Finding $\dfrac{dx}{dt}$ $\dfrac{dx}{dt}=\dfrac{2x}{y}$ Finding $\dfrac{dx}{dt}$ at $(2,9)$ The expression for $\dfrac{dx}{dt}$ depends on both the particle's $x$ -coordinate ${2}$ and its $y$ -coordinate ${9}$ : $\begin{aligned} \dfrac{dx}{dt}&=\dfrac{2({2})}{({9})} \\\\ &=\dfrac{4}{9} \end{aligned}$ In conclusion, the rate of change of the particle's $x$ -coordinate when the particle is at the point $(2,9)$ is $\dfrac{4}{9}$ units per minute.